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Explain solution RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 48 maths

Answers (1)

Answer:

\frac{-1}{3 \sin ^{3} t \cos 2 t}

Hint:

You must know the derivative of cos t and sin t function

Given:

\begin{aligned} &x=3 \cos t-2 \cos ^{3} t \\ &y=3 \sin t-2 \sin ^{3} t \quad \text { find } \frac{d^{2} y}{d x^{2}} \end{aligned}

Solution:

        \begin{aligned} &x=3 \cos t-2 \cos ^{3} t \\ &\frac{d x}{d t}=-3 \sin t+6 \cos ^{2} t \sin t \\ &=\sin t+\left(6 \cos ^{2} t-3\right) \\ &y=3 \sin t-2 \sin ^{3} t \\ &\frac{d y}{d t}=3 \cos t-6 \sin ^{2} t \cos t \\ &=\cos t\left(3-6 \sin ^{2} t\right) \end{aligned}

        \begin{aligned} &\therefore \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d x}{d t}=\frac{\cos t\left(3-6 \sin ^{2} t\right)}{\sin t+\left(6 \cos ^{2} t-3\right)} \\ &=\frac{\cos t\left(1-2 \sin ^{2} t\right)}{3\left(2 \cos ^{2} t-1\right)} \\ &=\frac{\cot t(\cos 2 t)}{\cos 2 t}=\cot t \\ &\frac{d y}{d x}=\cot t \\ &\frac{d^{2} y}{d x^{2}}=-cos e c^{2} t \end{aligned}

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